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Unformatted text preview: Chapter Twenty-Four ORGANIC CHEMISTRY
•
•
• Aliphatic Hydrocarbons
Aromatic Hydrocarbons
Functional Groups ALIPHATIC HYDROCARBONS
STUDY OBJECTIVES
1.
2.
3.
4. Name four different types of aliphatic hydrocarbons and draw a structural formula of a typical example of
each.
Name hydrocarbons of these four groups given a structural formula. Given a systematic name, draw a
structural formula.
Write equations for addition reactions to alkenes and alkynes.
Apply Markovnikov's rule concerning the addition of unsymmetrical reagents to alkenes. Organic Chemistry. The heart of organic chemistry is the carbon atom. Carbon is a key ingredient of
about six million chemical compounds primarily because of its ability to form long chains of self-linked atoms.
The existence of structural and geometric isomers contributes strongly to the number of organic or carboncontaining compounds.
The h ydrocarbons are an important class of organic compound that consist only of the elements carbon
and hydrogen. Hydrocarbons are divided into two classes: aliphatic and aromatic. A romatic hydrocarbons
contain one or more benzene rings. A liphatic hydrocarbons do not contain benzene rings. Four groups of
aliphatic hydrocarbons are known. These are alkanes, alkenes, alkynes, and cycloalkanes.
Alkanes. The general formula for an alkane is Cn H2n+2 , where n is the number of carbon atoms in the
molecule, n = 1, 2, 3…. When n = 1, we have the simplest member of the alkane family methane (CH4 ). The
alkanes make up a homologous series; a series of compounds differing in the number of carbon atoms. As n
increases one at a time, we can generate the formulas of the entire series of alkanes. The names and formulas of
the first ten straight-chain alkanes are given in Table 24.1. The first part of each name represents the number of
C atoms in the molecule. The ending '-ane' is common to all alkanes. If you learn these names, they will prove
to be very useful in naming other organic compounds.
Table 24.1 Names of the First Ten Alkanes.
_________________________
Formula
Name
_________________________
CH4
Methane
C 2 H6
Ethane
C 3 H8
Propane
C 4 H10
Butane
C 5 H12
Pentane
C 6 H14
Hexane
C 7 H16
Heptane
C 8 H18
Octane
C 9 H20
Nonane
C 10 H22 Decane
_________________________ 4 63
Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 64 / Organic Chemistry
Alkenes. Alkenes contain C C double bonds, and members of this homologous series have the general
formula Cn H2n . Alkenes are named with the same root word as alkanes to indicate the number of carbon atoms,
but all names end in "ene."
C 2 H4
C 3 H6
C 4 H8 CH2 CH2
CH3 CH CH2
CH3 CH2 CH CH2 Alkynes. Alkynes contain C
alkynes end with "yne."
C 2 H2
C 3 H4
C 4 H6 ethene (ethylene)
propene
butene C triple bonds, and have the general formula Cn H2n–2 . The names of CH CH
CH3 C CH
CH3 CH2 C CH ethyne (acetylene)
propyne
butyne Cycloalkanes. There is also a type of alkane that has atoms bonded into ring configurations. These are
the cycloalkanes. They have the same general formula as the alkenes, C n H2n+2 , but do not have double bonds.
Neither are these aromatics. Two cycloalkanes are shown below.
H2 C—CH 2
\/
CH2
cyclopropane H2 C—CH 2
|
|
H2 C—CH 2
cyclobutane Structural Isomers. Until now, we have considered only aliphatic hydrocarbons that have straight chains
of carbon atoms. Branching of hydrocarbon chains is very common. Butane (C4 H10 ) can be straight chained,
and branched chained.
CH3 —CH2 —CH2 —CH3 n-butane and CH3 —CH—CH3
|
CH3
2-methylpropane Note that both molecules have the same molecular formula, but they have different arrangements of atoms
(different structures). These are actually two distinguishable compounds with their different structures producing
slightly different chemical and physical properties. Molecules that have the same molecular formula, but a
different structure are called s tructural isomers . Straight chain hydrocarbons are called normal and use the
symbol n. The straight chain form of C4 H10 has the name n-butane. Naming branched chain hydrocarbons is
covered in the next section. In the alkane series, as the number of C atoms increases, the number of possible
structural isomers increases dramatically. For example, C4 H10 has 2 isomers, C6 H14 has 5, and C10 H22 has
75 isomers. Nomenclature. The rules for naming hydrocarbons according to the IUPAC system are briefly
summarized as follows:
1. 2. Back The systematic name of an alkane is based on the number of carbon atoms in the longest carbon chain. For
alkanes, the longest carbon chain is given a name corresponding to the alkane with the same number of C
atoms. The parent name of a compound with 5 carbon atoms in the longest carbon chain is pentane.
Groups attached to the main chain are called s ubstituent groups . Substituent groups that contain only
hydrogen and carbon atoms are called a lkyl groups . When an H atom is removed from an alkane the
fragment is called an alkyl group. This group can be attached to the longest chain. Alkyl groups are named
by dropping the ending -ane and adding yl to the alkane name. Therefore CH3 – is a methyl group, C2 H5 –
is an ethyl group, and C3 H7 is a propyl group. Table 24.2 in the text lists the names of six common alkyl
groups. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 65
3. 4. 5. 6. The locations of alkyl groups that are attached to the main chain must also be included in the name.
Number the carbon atoms in the longest chain. The numbering should start at the end of the chain such that
the side groups will have the smallest numbers.
Prefixes such as di, tri, and tetra are used when more than one substituent of the same kind is present.
Combine the names of substituent groups, their locations, and the parent chain into the hydrocarbon name.
Arrange the substituent groups alphabetically followed by the parent name of the main chain.
Alkanes can have many different types of substituents besides alkyl groups. Halogenated hydrocarbons
contain fluorine, chloroine, bromine and iodine atoms which are named fluoro, chloro, bromo, and iodo,
respectively. NH2 –, NO 2 –, and CH2
CH– are called amino, nitro, and vinyl, respectively.
When naming alkenes we indicate the positions of the carbon-carbon double bonds, and in alkynes the
position of the triple bond is numbered. For alkenes and alkynes, the parent name is derived from the
longest chain that contains the double or triple bond. Then parent name will end in ene for alkenes and yne
for alkynes. Number the C atoms in the main chain by starting at the end nearer to the double or triple
bond, and use the lower number of the carbon atom in the C C bond. These rules are applied in Examples 24.1 and 24.2. Addition Reactions. Alkenes, alkynes, and aromatics are called unsaturated hydrocarbons. This means
that they can acquire more hydrogen atoms in an addition reaction called hydrogenation.
CH2 CH2 + H2 → CH 3 —CH3
CH CH + 2H2 → CH 3 —CH3
Alkanes are called saturated hydrocarbons because they cannot acquire additional hydrogen atoms. The carbon
atoms in a saturated hydrocarbon are already bonded to the maximum number of H-atoms.
CH3 —CH3 + H2 → no reaction
In an addition reaction a small molecule such as H2 is added to an unsaturated hydrocarbon. The addition
reaction occurs at the C C double bond. One atom of the small molecule links to one of the carbon atoms of
the double bond, while the other atom attaches to the other carbon atom. Other examples are:
CH2
CH2
CH2 CH2 + C l 2 → CH 2 Cl—CH 2 Cl
CH2 + H2 O → CH 3 —CH2 OH
CH2 + H Cl → CH 3 —CH2 Cl Markovnikov's Rule. The two carbons of a double bond suffer different fates during addition reactions
with unsymmetrical reagents such as HCl, HBr, and H—OH (water). For example, two different compounds
might possibly be formed by reaction of 1-butene with HCl.
CH3 CH2 CH CH2 + H Cl → CH 3 CH2 CHCl—CH 3 or CH 3 CH2 CH2 —CH2 Cl
observed product
not found The rule that predicts which of the two products is formed is called M arkovnikov's rule . This rule states: In
the addition of unsymmetrical reagents to alkenes, the positive group of the reagent adds to the carbon that
already has the most hydrogen atoms. In HCl, HBr, and H2 O the hydrogen atoms have a partially positive
charge because they are bonded to more electronegative atoms. Thus, in the above reaction with
CH3 CH2 CH CH2 , the H atom from HCl bonds to the terminal C atom. The Cl atom bonds to the CH group. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 66 / Organic Chemistry
Geometric Isomers of Alkenes. C C double bonds are completely rigid. Consider the structures of cis- and trans-2-butene.
CH3
H
\
/
CC
/
\
H
CH 3
trans-2-butene CH3
CH3
\
/
CC
/
\
H
H
cis-2-butene Neither structure can rotate around the C C bond to become the other. These two molecules have different
physical properties and can be separated from one another. Therefore, these two structures represent two different
compounds and are isomers. Many pairs of such isomers have been isolated, and they have no tendency to
undergo interconversion.
The rigidity of the double bond gives rise to a new kind of isomerism: geometric isomerism , which is
sometimes called cis-trans isomerism. The relative positions of atoms in space is known as a configuration. The
isomers are named cis or trans depending on their configuration. The cis configuration has like groups on the
same side of the double bond.
CH 3 C H 3
\
/
–––C C–––
/
\
H
H
The trans configuation has like groups on the opposite sides of the double bond. Trans means across.
CH 3 H
\
/
–––C C–––
/
\
H
CH3
______________________________________________________________________
EXAMPLE 24.1 Naming Hydrocarbons
Name the molecule that has the following structure:
H 3 C CH 2 —CH3
|
|
CH3 —CH2 —C—CH—CH 2 —CH3
4|
3
2
1
5 CH 2
|
6 CH 2 —CH3
7 •Method of Solution
First identify the longest continuous carbon chain. Two equivalent chains containing seven carbon atoms are
noticeable. Number the carbon atoms as shown. Two ethyl groups appear at carbons 3 and 4, and one methyl
group appears at carbon 4. The parent name is heptane. Placing the names of the substituent groups
(alphabetically ethyl is before methyl) and their position numbers in front of the parent alkane name we get:
3,4-diethyl-4-methylheptane. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 67
•Comment
If the chain had been numbered in reverse order the name would be 4,5-diethyl-5-methylheptane.
Since 3,4 is smaller than 4,5 the first name is the correct one.
______________________________________________________________________
EXAMPLE 24.2 Naming Hydrocarbons
Name the following hydrocarbon:
6 5 4 CH3 —CH2 —CH—CH3
|
C H CH—CH3
3 2 1 •Method of Solution
Locate the longest chain that contains the double bond. This chain contains six C atoms so the root name will
be hexene. Number the chain so that the double bond has the smallest number, as shown. Note the methyl
group on carbon 4. The name is 4-methyl-2-hexene.
•Comment
The 2 indicates the position of the C C double bond, which connects carbons numbered 2 and 3.
_______________________________________________________________________
EXAMPLE 24.3 Addition Reactions
Give the structure of the product of the following reaction.
CH3 CH C—CH 3 + HCl
|
CH 3 •Method of Solution
This solution involves the addition of an unsymmetrical reagent to a double bond. According to Markovnikov's
rule the positive portion of the reagent (in this case an H atom) adds to the carbon atom in the double bond that
already has the most hydrogen atoms. The Cl atom adds to the other C atom. The product will be
Cl
|
CH3 CH2 —C—CH 3
|
CH3
____________________________________________________________________ EXERCISES
1.
2.
3. Write the general formulas of alkanes, alkenes, and alkynes.
Which of the classes of hydrocarbons in number 1 are unsaturated hydrocarbons?
Give the IUPAC name for the compound with the following structure.
CH 3
|
CH3 —CH2 —CH2 —CHCH2 —CH3 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 68 / Organic Chemistry
4. Give the IUPAC name for the following compound.
CH 3
|
CH—CH—CH 3 CH3 —CH 5. Name the following compounds.
a. CH 3 CHFCH 3
b. (CH 3 )3 CCH 2 CHClCH 3
6. Write the structure for 3,3-dimethyl pentane.
7. What is the product of the following addition reaction: Cl2 + CH3 CH
8. What is the product of the addition reaction HBr + CH3 CH CH2 ?
9. Namethe following: a. CH3 (CH2 )3 CH CHCH3 b . CH 3 (CH2 )3 C
10. Draw all the isomers of C 4 H8 . CH2 ?
CH3 AROMATIC HYDROCARBONS
STUDY OBJECTIVES
1.
2. Describe the structure of aromatic hydrocarbons.
Determine the product of a substitution reaction. Benzene. Benzene is the parent compound of a class of hydrocarbons called aromatic hydrocarbons. The
hydrogen to carbon ratio in the molecular formula of benzene C6 H6 suggests that it is highly unsaturated. In
1865 Kekule suggested that benzene had a cyclic structure with three double bonds.
H
H H H H
H Benzene can be represented as two resonance structures. Alternatively, benzene can be represented in terms of delocalized molecular orbitals. Regardless of which symbol is used, the hydrogen atoms are usually not explicitly written. Rather we must
remember that there is one attached to each carbon. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 69
Aromatic hydrocarbons contain benzene rings and do not have a simple general formula. Benzene is the
simplest compound in this group. Nomenclature of Aromatic Hydrocarbons. The naming of monosubstitited benzenes is
straightforward.
Br NO2 bromobenzene nitrobenzene If more than one substituent is present, we must indicate the position of the two groups relative to each other.
The prefixes ortho- (o), meta- (m), and para- (p) are used to indicate their relative positions.
Cl Cl Cl
Cl Cl
Cl
p-dichlorobenzene o-dichlorobenzenem-dichlorobenzene Reactions of Aromatic Compounds. Halogens react with benzene by a s ubstitution reaction
rather than an addition reaction. In this reaction an atom or group of atoms replaces an H atom in the benzene
ring. Chlorine atoms can be substituted for hydrogen atoms by reacting benzene and chlorine in the presence of
an FeCl 3 catalyst.
Cl + Cl2 FeCl 3
c atalys t + HCl EXERCISES
11. Name the following compounds.
Br
NO2 NO2
NO2
a. b. Br c. Br 12. How does benzene differ in reactivity from unsaturated aliphatic hydrocarbons? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 70 / Organic Chemistry FUNCTIONAL GROUPS
STUDY OBJECTIVES
1.
2. Define the term functional group. Sketch the general structural formulas of alcohol, ether, ketone, aldehyde,
carboxyl, ester, and amine groups.
Write structural formulas for products of the oxidation of alcohols and aldehydes, for products of the
reactions of carboxylic acids with alcohols, and for products of the reactions of esters with water and sodium
hydroxide. Alcohols. Alkanes are quite inert toward most substances, the main exception being oxygen. The alkane
portion of organic compounds is combustable. Certain groups of atoms within carbon compouinds always react
in the same way regardless of the chain length of the alkane portion. The group of atoms that is largely
responsible for the chemical behavior of a molecule is called a f unctional group . For instance, alcohols
contain an alkane group and a hydroxyl group, OH. Methanol (CH3 OH), ethanol (C2 H5 OH), and n-propanol
(C3 H7 OH) are the first three alcohols of a homologous series. In each of the three, chemical reactions occur at
the OH group, rather than at the less reactive alkane group. The hydroxyl group is the functional group in each,
and the three alcohols react toward other reagents in the same way.
Reactions of alcohols discussed in the textbook are oxidation by oxidizing agents, displacement of hydrogen
by alkali metals, and esterification. An oxidizing agent removes two H atoms, one from the hydroxyl group and
one from the adjacent carbon atom, forming an aldehyde (shown below). Permanganate ion is an oxidizing
agent.
– MnO4 CH3 CH2 CH2 –OH → CH3 CH2 CHO
Quite often, in organic reactions, we do not write completely balanced equations. Here the emphasis is on the
chemical change of the alcohol functional group.
Alkali metals displace H2 (g) from alcohols. Potassium is a reducing agent.
2CH3 CH2 OH + 2K → 2CH 3 CH2 OK + H 2
Esterification will be discussed under carboxylic acids. Ethers. The functional group in ethers is C—O—C or, more generally, R—O—R, where R stands for
any alkyl group such as those listed in Table 24.2 of the textbook. Ethers and alcohols are structural isomers.
As illustrated in the following diagram, both dimethyl ether and ethanol have the same chemical formula.
CH3 OCH3
dimethyl ether
Tb –25°C ← C 2 H6 O → CH3 CH2 OH
ethanol
Tb 78°C The boiling points given show that ethers are much more volatile than the corresponding alcohols. This is
because of the absence of hydrogen bonding in ethers. Alcohols contain the polar hydroxyl group, Oδ – –Hδ +,
which can participate in hydrogen bonding to neighboring alcohol molecules.
For many years diethyl ether, CH3 CH2 OCH2 CH3 , was used as an anesthetic. Aldehydes and Ketones. The funtional group in these compounds is the carbonyl group, C
aldehydes the carbonyl group is at the end of the alkane chain. In ketones it is not a terminal group.
O
||
CH3 CH2 CH2 CH or CH3 CH2 CH2 CHO
butanal (an aldehyde) Back Forward Main Menu TOC O. In O
||
CH2 CH2 CCH 3 or CH3 CH2 COCH3
2-butanone (a ketone) Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 71
Aldehydes are easily oxidized to carboxylic acids.
O
O
–
| | MnO4
||
CH3 CH2 CH → C H 3 CH2 C—OH
Ketones generally are less reactive than aldehydes.
O
–
||
MnO4
CH3 CH2 CCH 3 → no reaction Carboxylic Acids. The carboxyl group is the functional group in organic acids.
O
||
—C—OH abbreviated as —COOH
This group is weakly acidic. The ionizable hydrogen atom accounts for the acidic properties of carboxylic acids.
CH3 COOH(aq) CH 3 COO– (aq) + H+(aq) A homologous series of organic acids can be generated starting with formic acid.
HCOOH
CH3 COOH
CH 3 CH2 COOH
CH3 CH2 CH2 COOH methanoic acid (formic acid)
ethanoic acid (acetic acid)
propanoic acid
butanoic acid Fatty acids are carboxylic acids that contain more than four carbon atoms. Two fatty acids are:
CH3 (CH2 )14 COOH
CH3 (CH2 )16 COOH palmitic acid
stearic acid Esters. The ester functional group is
O
||
—C—OR or —COOR
where R stands for any alkyl group. This functional group resembles the carboxylic acid group except that the R
group replaces the H atom. Carboxylic acids react with alcohols to form the compounds called esters. The
reaction that changes an acid to an ester is called an e sterification reaction.
O
O
||
||
CH3 C— OH + H —OCH3 → CH3 C—OCH3 + H 2 O
acetic acid
methanol
methyl acetate
Esterification is an example of a condensation reaction. A "condensation" is a reaction in which parts of two
molecules become joined by formation of a new covalent bond to make a new, larger molecule. Esterification
reactions are reversible, and at equilibrium a mixture is formed that contains all four substances.
Saponification is the alkaline hydrolysis of an ester. In this reaction the base reacts directly with an ester to
split it into a salt of an organic acid and an alcohol.
C 2 H5 COOCH3 + NaOH → C 2 H5 COONa + CH 3 OH Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 72 / Organic Chemistry
Amines. Organic bases contain the —NH2 functional group, which is called an amino group. Molecules
containing an amino group are called amines. The amines may be considered as derivatives of ammonia (NH3 ).
Amines have the general formula R3 N, where R may be H, an alkyl group, or an aromatic group. Molecules of
the type RNH 2 are called primary amines. Those with R2 NH are secondary amines, and R3 N is a tertiary amine.
H
|
CH3 —N—H
methylamine
a primary amine CH 3
|
CH3 —N—H
dimethylamine
a secondary amine CH3
|
CH3 —N—CH3
trimethylamine
a tertiary amine ____________________________________________________________________
EXAMPLE 24.4 Identifying Functional Groups
Indicate the functional groups in the following formulas:
a. C 5 H11 OH
b. CH3 CHO
c. C 3 H7 OCH3
d. CH3 COC 2 H5
e. CH3 COOCH3
•Method of Solution
Learning to recognize functional groups requires memorization of their structural formulas. Table 24.5
(textbook) shows a number of the important functional groups.
a. C 5 H11 OH contains a hydroxyl group. It is an alcohol. b.
c.
d.
e. O
||
CH3 CHO is a way to represent CH3 C—H on a single line of type.
C O is a carbonyl group. CH3 CHO is an aldehyde.
C 3 H7 OCH3 contains a C—O—C group and is an ether.
CH3 COC 2 H5 is a way to represent a ketone on a single line. C O is a carbonyl group.
CH3 COOCH3 is a condensed structural formula for:
O
||
CH3 —C—O—CH3 CH3 COOCH3 is an ester functional group.
________________________________________________________________________
EXAMPLE 24.5 Predicting Reaction Products
Predict the product or products of the following reactions:
2–
a. CH3 CH(OH)CH3 + Cr2 O7 →
–
b. CH3 CHO + MnO4 →
c. HCOOH + CH 3 OH →
d. CH3 CH2 COOCH3 + NaOH →
•Method of Solution
a. This is the reaction of a secondary alcohol with an oxidizing agent. The product is a ketone:
O
||
CH3 —C—CH 3 . Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 73
b.
c. This is the reaction of an aldehyde with an oxidizing agent. The product is a carboxylic acid: CH3 COOH.
The reaction of an alcohol with an acid yields an ester and water.
O
||
HCOCH3 + H2 O d. This is a saponificaton of an ester to form the sodium salt of a carboxylic acid and an alcohol. O
||
CH3 CH2 CONa + CH 3 OH
________________________________________________________________________ EXERCISES
13.
14.
15.
16.
17.
18.
19. Draw structures for the following functional groups: aldehyde, amine, carboxylic acid, ester.
Give two chemical properties of alcohols.
Oxidation of what type of compound yields a ketone?
What elements are present in an amine group?
Which of the following functional groups exhibit hydrogen bonding? carboxylic acids, alcohols, ethers.
What are the reactants in a saponificaton reaction that yields C6 H13 COONa and C2 H5 OH?
Identify the functional group or type of molecule in the following.
a. CH 3 OH
b. CH3 CHO
c. CH 3 COCH3
d. CH3 COOCH3
e. CH 3 CH2 CH3
_______________________________________________________________________________ PRACTICE TEST
1. Give the systematic name for each of the following structural formulas.
a. CH3 (CH2 )7 CH3
b. CH 3
|
CH3 —CH2 —CH—CH CH 2 c. d. CH3 —CH2 —C e. Back CH3 —CH2 —CH—CH2 —OH
|
CH 2 —CH3 CH 2 —CH3
|
CH3 —CH2 —CH—CH— CH—CH 3
|
|
CH2 —CH2 CH3
|
CH3 Forward Main Menu C—CH 2 —CH—CH3
|
C 3 H7 TOC Study Guide TOC Textbook Website MHHE Website 4 74 / Organic Chemistry
f. CH 3
|
C H—CH 2 —CH3
|
CH3 —CH2 —CH
|
CH 2 —CH2 —CH3 g. H
\
C
/
CH3 CH 2 CH3
/
C
\
H h. i.
2. 3.
4. 5. CH3 —CH—CH2 —CH3
|
OH
CH3 —OH Draw structural formulas for:
a. 3-ethyl-4-methyl-4-isopropylheptane
b. 3-bromo-2,5-dimethyl-trans-3-hexene
c. 3-methyl-3-hexanol
d. trichloroacetic acid
Draw structural formulas of all isomers of C 5 H10 . Include both structural and geometric isomers.
List the functional groups by name that are shown in the following molecules.
a. CH3 –CH CH–CH–CH 2 –NH2 b .
O
O
O
|
||
||
||
OH
HO–C–CH2 –C–CH 2 –CH2 –CH
Give the structure of the organic product of each of the following reactions:
a.
OH
|
CH 3 CH2 CHCH3
agent
b. O
||
CH3 CH agent CH + Cl2 → c. CH3 C d. C 3 H8 + C l 2 + H2 light Pt
catalys t e.
f. CH3 CH2 CH CHCH 2 CH3 + H2 O g. 2–
CH3 CHCH3 + Cr2 O7 dil → |
OH h. Back O
||
2–
CH3 CH2 CCH 3 + Cr2 O7 Forward Main Menu → TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 75 + H2O i.
j. O
||
CH3 COH + CH3 OH k. CH3 CH2 CH
6. a. H 2 SO4
dil H+ CHCH 3 + H2 Sketch structural formulas of the hydrolysis products of the following ester:
CH 3 O
|
||
CH3 —CH—C—CH 3 + H 2 O b. Pt H+ Sketch structures of the products of the following saponification reaction:
CH3 (CH2 )12 COO(CH2 )4 CH3 + NaOH → 7. A certain reagent converts methanol into methanoic acid. In this reaction the alcohol reacts as:
a. a Lewis base b. a Lewis acid c. a catalyst d. an oxidizing agent e. a reducing agent.
8. Define the following terms:
a. an alkene b. homologous series c. an addition reaction d. functional group.
9. How many structural isomers are there of C4 H10 O?
10. Show how the following chemical change can be carried out in two steps.
CH3 CH2 CH2 OH → CH3 CHCH3
|
OH ANSWERS Exercises
1.
2.
3.
4.
5.
6. Back C n H2n+2 , C n H2n , and C n H2n–2
alkenes and alkynes
3-methylhexane
4-methyl-2-pentene
a. 2-fluoropropane b. 2-chloro-4,4-dimethylpentane
CH 3
|
CH3 —CH2 —C—CH 2 —CH3
|
CH 3 Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 4 76 / Organic Chemistry
7.
8.
9.
10.
11.
12.
13. 14.
15.
16.
17.
18.
19. CH3 CHCl—CH 2 Cl
CH3 CHBr—CH3
a. 2-heptene b. 1-hexyne
Six isomers. 1–Butene, cis- and trans-2-butene, 2-methyl propene and 2 cycloalkanes.
a. o-dinitrobenzene b. p-dibromobenzene c. m-bromonitrobenzene
Benzene is much less reactive than alkenes. Rather than addition reactions, benzene undergoes substitution
reactions.
O
H
O
O
|| |
||
||
—C—H
—N—H—C—O—H
—C—O—R
Alcohols react with carboxylic acids to form esters. Also alcohols are easily oxidized.
A secondary alcohol
Nitrogen and hydrogen
Carboxylic acids and alcohols
An ester; C 6 H13 COOC2 H5 and NaOH.
a. alcohol b. aldehyde c. ketone d. ester e. alkane Practice Test
1. n-nonane
b. 3-methyl-1-pentene
6-methyl-3-nonyne
e. 3,4-diethyl-2-methylheptane
trans-2-pentene
h. 2-butanol
C 2 H5 CH3
|
|
CH3 —CH2 —CH—C—CH 2 —CH2 —CH3
|
C H 3 —CH—CH3 b. 2. a.
d.
g.
a. CH 3
||
CH3 —CH—C
|
Br H
C—CH—CH 3
|
CH 3 c. 4. Back CH 3
|
CH3 —CH2 —C—CH 2 —CH2 —CH3
|
OH d.
3. c. 2-ethyl-1-butanol
f. 4-ethyl-3-methylheptane
i. methanol CCl 3 –COOH There are five alkenes: 1-pentene, 3-methyl-1-butene, 2-methyl-1-butene, and cis- and trans--2-pentene.
There are three cycloalkanes: cyclopentane, methylcyclobutane, and dimethylcyclopropane.
a. Carbon-carbon double bond, hydroxyl, amine
b. Carboxyl, carbonyl (ketone), carbonyl (aldehyde) Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Organic Chemistry / 4 77
5. a. b.
c.
d. O
||
CH3 CH2 CCH 3
CH3 COOH
CH3 CCl 2 CHCl 2
C 3 H7 Cl, C 3 H6 Cl 2 , C 3 H5 Cl 3 ,etc. e. f. g. CH3 CH2 CH2 CHCH2 CH3
|
OH
O
||
CH3 CCH 3 h. no reaction i.
OH
j. k.
6. a. O
||
CH3 COCH3
CH3 CH2 CH2 CH2 CH3
CH 3 O
|
||
CH3 —CH—C—OH + CH3 OH b. CH3 (CH2 )12 COONa + CH 3 (CH2 )4 OH 7. b.
8. a.
b.
c. A straight chain or branch chain hydrocarbon with one or more double bonds
Organic compounds with the same functional group that differ only in the number of carbon atoms.
A reaction in which a small molecule is added to an organic compound. The organic molecule must
contain a C
C bond or C O (carbonyl) group.
d. A characteristic grouping of atoms that imparts certain chemical and physical properties when
incorporate into organic molecules
9. Seven (four alcohols and three ethers) 10. CH3 CH2 CH2 OH
CH3 CH CH3 CH conc CH 2 + H 2 O dil CH 2 + H2 O CH 3 CHCH3
|
OH _______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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